Analog circuits Test 2

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Analog circuits Test 2

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Weekly Quiz Competition

Question 1

2 / -0.33

In CE configuration, effect of bypass capacitor is/are ______

Voltage Gain increases

Bandwidth increases

Input resistance decreases

DC collector current increases

Solution

CE with a bypass capacitor	CE capacitor without a bypass capacitor
Voltage Gain : A_v = -g_m R_L’	\({A_v} = \frac{{ - {g_m}R_L'}}{{1 + {g_m}{R_E}}}\)
Input Impedance : R_i = r_π	R_i = r_π + (1 + β) R_E
Bandwidth decreases	Bandwidth increases
DC collector current increases	DC collector current decreases.

Question 2
2 / -0.33

A Half wave rectifier uses a capacitor of 680 μF in parallel to a load of 2kΩ. The average output voltage is 40 V and the line frequency is 60 Hz.
Which of the following is/are true?

A
Peak to Peak ripple (V_s) = 0.49 V

B
Percentage duty cycle of diode = 2.48%

C
Capacitor filter will result in smaller ripple factor when R_L is large

D
Peak to Peak ripple voltage increases with frequency.

Solution

Formula used:
V_DC = I_DCR_L
Ripple voltage is calculated as:
\({V_r} = \frac{{{I_{DC}}}}{{{f_o}C}}\)
Ripple factor, r is given by:
\(r = \frac{1}{{2\sqrt 3 {f_o}C{R_L}}}\)
Also, the angle of conduction is:
\(= \sqrt {\frac{{2{V_r}}}{{{V_m}}}} \)
\({V_m} = {V_{DC}} + \frac{{{V_r}}}{2}\)
Calculation:
Given: V_DC = 40 V, R_L = 2 kΩ, f_o = 60 Hz, and C = 680 μf
\({I_{DC}} = \frac{{{V_{DC}}}}{{{R_L}}} = \frac{{40}}{2}mA\)
I_DC = 20 mA
Now, the ripple voltage will be”
\({V_r} = \frac{{{I_{DC}}}}{{{f_o}C}} = \frac{{20 \times {{10}^{ - 3}} \times {{10}^6}}}{{60 \times 680}}\)
V_r = 0.49 V
\({V_m} = {V_{DC}} + \frac{{{V_r}}}{2} = 40 + \frac{{0.49}}{2}\)
V_m = 40.24 V
The angle of conduction will now be:
\(\sqrt {\frac{{2{V_r}}}{{{V_m}}}} = \sqrt {\frac{{2 \times 0.49}}{{40.24}}} \)
= 0.156 radian
Duty cycle \(= \frac{{angle\;of\;conduction}}{{2\pi }} \times 100\%\)
\(= \frac{{0.156}}{{2\pi }} \times 100\%\)
= 2.48%
Options 1) & 2) are correct
The Ripple factor for capacitor filler will be:
\(r = \frac{1}{{2\sqrt 3 {f_o}C{R_L}}}\)
\(r \propto \frac{1}{{{R_L}}}\)
For large R_L, the ripple factor will be smaller.
Also, since the Peak to peak ripple voltage is:
\({V_r} = \frac{{{I_{DC}}}}{{{f_o}C}}\)
\({V_r} \propto \frac{1}{{{f_o}}}\)
Peak to peak ripple voltage decreases with frequency.
Options 1, 2 & 3 are correct.